Particle Art

Bubble chamber shows the path made by particles. The particles are not known. Need further investigation. Spiraling ones shows the particle losing speed....
Identify the start and end of the path. Have fun.

Particle tracks in cloud chamber

cloud chamber and B field applied Alpha tracks. The alpha particles have high ionising power and approx the same energy.
In a cloud chamber or bubble chamber , tracks of alpha, beta and gamma can be seen as the ions formed along its tracks can be made visible by producing condensation of vapour just like the trails of an aeroplane in the sky. The ionisation power of gamma is weak and therefore tracks from gamma are difficult to see.
The particles curve is opposite direction due to formation of opposite charges subjected to magnetic field. The tracks need proper identificaiton to know what produces it.

Binding energy and stablity of atom

Binding energy is the energy that is lost when a nucleus is created from protons and neutrons. If you added up the total mass of the nucleons (protons and neutrons) that compose an atom, you would notice that this sum is less than the actual mass of the atom. This missing mass, called the mass defect, is a measure of the atom's binding energy. It is released during the formation of a nucleus from the composing nucleons. This energy would have to be put back into the nucleus in order to decompose it into its individual nucleons. The greater the binding energy per nucleon in the atom, the greater the atom's stability....excerpts from libraryquest.com

More elementary particles

An antiproton (blue) enters a bubble chamber from bottom left and strikes a proton. The released energy creates four positive pions (red) and four negative pions (green). The yellow streak at the far right is a muon, a decay product of the adjacent pion. (The dark blue curlicues are low-energy electrons knocked from atoms, not involved with the antiproton.)pic taken from www.lbl.gov/Science-

Release of energy results in stability of atoms

Achieving greater stability results in release of energy. pic taken from ec.europa.eu
Fusion : releasing energy when two small nucleus combine to form larger nucleus
Fission: releasing energy when a very large nucleus breaks into smaller nucleus

Binding energy

Energy needed to separate the nucleus into into its constituents : nucleons . pic taken from sol.sci.uop.edu/watch web: phys.unsw.edu

Three types of radiation

borrowed from faculty.weber.edu
borrowed from freedomforfission.org

Rutherford's discovery

If you fire bullets at a piece of tissue paper , how can it rebound and hit you!
Read link here
See the nucleus to visualise the force involved in deflecting alpha to a large extent.
pic borrowed from newcastle-school.org

What's the difference between Thomson's plum pudding model and Rutherford's model

The large force observed in the alpha-scattering cannot be explained using the plum pudding modelwhere the charges are scattered inside the atom. The scattered charges in plum pudding does not produce large deflections as observed. The forces measured were also too large which suggest that the alpha approach very near the high positive charge deep inside the 'atom'.
See animation hereSee thomson's and rutherford's.

Plum pudding model versus Rutherford's model

The electrons are blue and the positive charges are red (as shown).The large force can only arise if alpha had approached deeper into the atom which also imply that the nucleus is concentrated at the centre and that it has lots of empty space.
See Plum pudding model vs Rutherford's model

Lasers to enable fusion

High power laser could initiate fusion reported in BBC : link to delicious

What is the mass of electron travelling at 0.9c?

To find the mass of the electron :
Calculate the KE = o.5mass of electron x(o.9c)2 equate to mc2.
m= 0.5mass of electron x 0.81= 0.405 mass of electron.
The mass of electron = rest mass + 0.405 rest mass = 1.405 rest mass
The mass of electron moving near speed of light is actually about 1.405 times.

Energy released or required?

CONSERVATION OF MASS ENERGY - MASS DIFFERENCE

To calculate energy released or required in a nuclear reaction of any kind, we simply add all masses before and after.

Mass Difference = Total mass "before" - Total mass "after"

We can then simply convert to MeV.

Consider the nuclear collision when an alpha particle collides with N-14.The product is hydrogen and oxygen. The mass on the LHS is smaller than the RHS. Therefore energy input is required. The energy is carried by a fast alpha particle.

Mass Before Mass After
He 4.00260 H 1.00783
N 14.00307 O 16.99913
Total 18.00567 18.00696

In this case the mass AFTER reaction is greater than before! The alpha particle MUST have adequate kinetic energy for this reaction to proceed!

Mass Difference = 18.00567 - 18.00696 = - 0.00129 u

Energy required = 931 x 0.00129 MeV = 1.20 MeV
extracted from tased.com

What is rest mass?

A atom at rest is said to have rest mass. In the data provided , the mass is always rest mass. An atom released in a nuclear reaction is not at rest but carries with it kinetic energy . An atom with energy is said to have a larger mass than its rest mass due to einstein's mass-energy relation, E = mc2.

Conservation of mass and energy

An reaction is possible if the RHS and LHS satisfy conservation of mass-energy.
Example :
A = B + C (But mass of A is larger than B+C) For this reaction to occur , the energy released account for the difference in mass .
If the mass of A is smaller than B+C , than the reaction will not occur. To initiate the reaction would require A to possess some energy equivalent to the difference in the mass. This energy may be in the form of KE or absorption of some high energy photon.

Rutherford's scattering experiment

When helium nucleus is used to hit gold target. These observations allow rutherford to make the following conclusions.
1. Most of the helium nucleus pass through gold film undeflected.
The atom is mostly empty space.
2. A small percentage of helium nucleus are deflected by large angles.
The atom has positive and massive nucleus concentrated at a small point. A large couloumb force can explain the large deflection produced on the alpha particles.
See animated motion of alpha particle by michael fowler.
Another animation from uni of colorado

Binding energy

Binding energy can be understood as energy given off when a nucleus is formed from nucleons . It can also be considered as energy need to break off the nucleus into its constituent nucleons. If A = B + C + 2 neutrons. You can start with all the nucleons you need for forming A . Now energy is released to make nucleus A which is BE (A) . Starting with the same nucleons you can also make B and C which is BE(B) + BE(C). Since BE on the right hand side is more than the BE on the left hand side , more energy is released in forming B and C from A . Hence the difference in the BE is the energy released.

Energy released in a reaction

A = B + C + 2neutrons
The equation is balanced i.e no of nucleons are the same on both sides of the equation.
To find the energy released using binding energy ignore the neutrons as they do not have binding energy as they are single neutrons .
Find energy needed to break up A, B and C into its constituents i.e nucleons
You will find that the energy to break up A is lower that the energy needed to break up B and C together , as B and C are more stable products than A.
Therefore in forming B and C from A there will be further release of energy. This energy can be calculated by BE(B) + BE(C) - BE (A)

Nuclei or atom

To find the binding energy of a nucleus , it does not involve the electrons . If an atom mass is given it is inclusive of all the electron mass. If nuclei mass is given it does not include electrons. Therefore it is only logicial to take away electrons mass from the atom mass so that the nuclei mass can be obtained. Example: Binding energy of C is [12x mass of proton + 12 x mass of neutron - mass of C nuclei] x c*2.The mass of C atom is 12u . Therefore the mass of C nuclei is
12 u -12 x mass of electron.

one atomic mass unit gives 931 Mev

1 a.m.u or 1 u is 1.66x 10*-27kg
Using E=mc*2 , the energy turn out to be 1.66 x 10*-27kg x square of speed of light.
approximately 931 MeV(rough value not to be used for examination purpose)